Android:按钮onclick显示错误

Question

对于我的两个按钮,我想实现一个开关盒,点击后我得到以下错误如何解决这个问题？

void是变量onClick上的无效类型令牌")"上的语法错误; 预期

这是我的代码.

        Button regmalebtn = (Button) findViewById(R.id.regmalebtn);
        Button regfemalebtn = (Button) findViewById(R.id.regfemalebtn);

    String gender; 
    gender = regmalebtn.getText().toString();
  public void onClick(View v) {

      switch(v.getId()){

      case R.id.regmalebtn:
    //    String gender; 
    //    gender = regmalebtn.getText().toString();
        gender.equals("M");
    //  request.addProperty("gender",gender );

           break;

      case R.id.regfemalebtn:
          gender = regfemalebtn.getText().toString();
            gender.equals("F");
            //request.addProperty("gender", gender);
           break;
           default:
               break;
      }
  }  

Answer 1

您收到以下错误

令牌上的语法错误")",; 预期

That means onClick() method is defined inside another method. It should be defined outside of any method in the same class

谢谢迪帕克